Consider the curve given by the equation xy - 3x + 2y = -2. Find the equation of the tangent in the point (2,1).
So through implicit differentiation I've found that (if I'm correct) y' = (1-y)/(x-2)
How do I proceed to find the tangent line since there are still 2 variables in the function?
Note that if we have $$xy -3x+2y=-2$$ then after differentiation we get $$y+xy' - 3 +2y'=0 \implies xy'+2y'=3-y \implies y'(x+2)=3-y \\ \implies y'=\frac{3-y}{x+2}$$ Edit: So to find the tangent line, first you need the slope of the tangent line at the point $(2,1)$. You do that by plugging in the coordinates in the derivative expression we found, so $y'$ at $(2,1)$ is: $$y'=\frac{3-1}{2+2}=\frac12$$ You know the slope, and you know one point on the tangent line (namely, $(2,1)$) so you can find the equation of the line from here.